Surface Area and Force Questions

[Flash]

Okay, if you've seen my last two topics, you'll have noticed that I need to brush up on some physics and clear some of my misconceptions. So I need to ask some questions.



The above picture shows the cycling of a High Efficiency Shocker bolt. I noticed that the bolt rod is curved in a lot of places, especially around the o-ring. Does anyone know why? Is it just for structural integrity or does it serve some other purpose?

Here's the next picture.



Its the diagram of the QEV from the wiki. What function do the conical shapes on the sealing and pilot sides provide? Basically, why is it shaped the way it is? (in terms of surface area and its relation to force)

[Ragnarok]

I'm not sure what you mean about the bolt, but for the QEV...

Those conical shapes have no effect on the opening force.
Now, the higher area might seem to suggest otherwise, but as pressure always acts normal (or perpendicular, which ever term you rather, but normal is the term used in engineering for it), only the component of the force actually along the axis of travel will actually count.

The higher area and component of force perfectly cancel each other.
So, the shape doesn't improve opening time. As for why it's shaped that way, I haven't the faintest.

The QEVs I own don't have those shapes, so I can't assume the reason is that much of a concern, or surely mine would echo the animation.

[pizlo]

My best bet for the QEV is for an easy sealing face, as for the bolt thingy, the curves around the o Oring don't do anything.

[clide]

The QEV piston is shaped like that so it acts as a one way valve and lets air flow through it when pressure is higher on the pilot side, but it gets pressed up against the wall and seals off when pressure is higher on the chamber side. The smaller cones are probably either for improved flow or piston stability.

I'm not really sure about the bolt and what curves I should be looking at, but it is probably a flow and/or strength/weight thing.

[Flash]

The gray rod inside of the bolt is what I called the bolt rod.

Now, the higher area might seem to suggest otherwise, but as pressure always acts normal (or perpendicular, which ever term you rather, but normal is the term used in engineering for it), only the component of the force actually along the axis of travel will actually count.

So basically, only the flat surfaces facing the direction it's moving are the only forces that contribute force? And what do you mean by perpendicular?

Thanks for all the info guys.

[Daegurth]

perpendicular, or normal, means at right angles to the surface- that is, sticking straight out of the surface, and all angles between the normal line and the surface are 90 degrees. for further reading on this, look quickly at refraction through blocks in optics- there is always a normal line drawn, and you can see it's attitude in comparison to the surface of the glass/refractive medium.

and no, any other surfaces that aren't parallel to the direction of travel will contribute some force, you'll have to resolve them into vectors- there's plenty of information about that on the internet though, just google it.

[jimmy101]

So basically, only the flat surfaces facing the direction it's moving are the only forces that contribute force? And what do you mean by perpendicular?

Nope. Only the projection along the direction (axis) of travel of the surface.

The easiest way to look at it (without resulting to vector analysis or Calculas) is that the area is simply the area of the shadow the object casts when iluminated along the axis of travel.

Or, even simpler, for a projectile that is a tight fit in a barrel the area is just the area of the barrel. It doesn't matter what the actual shape of the surface is as long as it's cross section is the same as the barrel. A shell with a flat butt, or a concave, or a pointy, or ... all have the same cross sectional area and all will experience the same force due to pressure along the only axis you really care about, the axis of travel.